Ta có 

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)

\(\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\)

\(\Rightarrow ayz+bxz+cxy=0\)

Ta có

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)

\(\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=1\)

\(\Rightarrow\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

\(\Rightarrow\frac{2xy.abc^2+2yz.a^2bc+2xz.ab^2c}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

\(\Rightarrow\frac{2abc.\left(cxy+ayz+bxz\right)}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

 Ta có \(cxy+ayz+bxz=0\)

\(\Rightarrow\frac{2abc.\left(cxy+ayz+bxz\right)}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

\(\Rightarrow\frac{2abc.0}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

\(\Rightarrow1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)=0\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đpcm\right)\)

bài này bạn bình phương vế thứ 2 lên rồi phân k vế 1 là ra đấy

\(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\left(x,y,z\ne0\right)\).

Ta có:

\(a+b+c=0\).

Ta phải chứng minh rằng nếu \(a+b+c=0\)thì \(a^3+b^3+c^3=3abc\).

Thật vậy, xét hiệu  \(A=a^3+b^3+c^3-3abc\)với \(a+b+c=0\).

\(A=\left(a+b\right)^3-3ab\left(a+b\right)-3abc+c^3\).

\(A=\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]\).

\(A=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\)\(-3ab\left(a+b+c\right)\).

\(A=\left(a+b+c\right)\left(a^2+2ab+b^2-ab-ac+c^2-3ab\right)\).

\(A=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\).

\(A=0\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)(vì \(a+b+c=0\)).

Do đó \(a^3+b^3+c^3-3abc=0\).

\(\Rightarrow a^3+b^3+c^3=3abc\)với \(a+b+c=0\)

\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)với \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)(điều phải chứng minh).

\(\frac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)

\(\Rightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)

\(\Rightarrow a^2x^2+b^2y^2+c^2z^2+2abxy+2acxz+2bcyz\)\(=a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)

\(\Rightarrow b^2x^2-2abxy+a^2y^2+b^2z^2-2bcyz+c^2y^2+a^2z^2-2acxz+c^2x^2=0\)

\(\Rightarrow\left(bx-ay\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}bx-ay=0\\bz-cy=0\\az-cx=0\end{cases}\Rightarrow\hept{\begin{cases}bx=ay\\bz=cy\\az=cx\end{cases}\Rightarrow}\hept{\begin{cases}\frac{b}{y}=\frac{a}{x}\\\frac{b}{y}=\frac{c}{z}\\\frac{a}{x}=\frac{c}{z}\end{cases}\Rightarrow}\frac{a}{x}=\frac{b}{y}=\frac{c}{z}}\)

Ta có: x²=yz (1)

         y²=xz (2)

         z²=xy  (3)

Cộng từng vế các BĐT (1);(2);(3) ta được:

x²+y²+z²=yz+xz+xy

<=>2(x²+y²+z²)=2(yz+xz+xy) (nhân cả 2 vế cho 2)

<=>2x²+2y²+2z²=2yz+2xz+2xy

<=>(2x²+2y²+2z²)-(2yz+2xz+2xy)=0

<=>2x²+2y²+2z²-2yz-2xz-2xy=0

<=>(2x²-2xy)+(2y²-2yz)+(2z²-2xz)=0

<=>(x-y)²+(y-z)²+(z-x)²=0

Vì \(\left(x-y\right)^2\ge0\)   với mọi x;y

\(\left(y-z\right)^2\ge0\)   với mọi y;z

\(\left(z-x\right)^2\ge0\) với mọi z;x

=>(x-y)²+(y-z)²+(z-x)² \(\ge\) 0 với mọi x;y;z

Theo đề: (x-y)²+(y-z)²+(z-x)²=0

=>(x-y)²=(y-z)²=(z-x)²=0

<=>x-y=y-z=z-x=0

+)x-y=0=>x=y (4)

+)y-z=0=>y=z (5)

+)z-x=0=>z=x (6)

từ (4);(5);(6)=>x=y=z (ĐPCM)

Ta có: x²=yz =>\(\frac{x}{y}=\frac{z}{x}\) (1)

y²=xz => \(\frac{x}{y}=\frac{y}{z}\) (2)

Từ (1);(2) =>\(\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta được:

\(\frac{x}{y}=\frac{z}{x}=\frac{y}{z}=\frac{x+z+y}{y+x+z}=1\)

Do đó, x=y*1=y

          z=x*1=x

=>x=y=z

Vậy x=y=z